EITFastTrack.com (EBook and Practice Exams) Forum
General Category => Problem of the Day !!! => Topic started by: FastTrack Guru on November 17, 2014, 08:05:45 AM

Question:
An applied force of 150 pounds acts on a helical spring with mean spring diameter of .25 inches. Find the shear stress (lb/in2) if the wire diameter is .08 inch.
(A) 1879
(B) 2100
(C) 3812
(D) 4103

answer: c

Refer to page 224 of the FE Handbook: Helical Springs
The equation of shear stress is:
T = Ks (8FD)/ πd3
You just have to calculate Ks:
Ks= (2C + 1)/(2C)
C = D/d
C= .25 in/ .08 in = 3.124
Ks= (2(3.124) + 1)/(2(3.124))
= (7.248)/6.248
= .022
T = Ks (8FD)/ πd3
T = .022(8)150 lb (.25in) / π(.08 in )3
T = 6.6 in lb / .00160845 in3
T = 4103 lb / in2
Answer is (D) 4103 lb / in2

Is Ks actually .022?

It appears that Ks should be 1.16:
Ks= (2C + 1)/(2C)
C = D/d
C= .25 in/ .08 in = 3.124
Ks= (2(3.124) + 1)/(2(3.124))
= (7.248)/6.248
= .022 (!?)
= 1.16

Using Ks = 1.16, I come up with T = 216351 lb/in2